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Arte-miy333 [17]
2 years ago
7

How to multiply complex fractions​

Mathematics
1 answer:
ASHA 777 [7]2 years ago
7 0

Step-by-step explanation:

Create a single fraction in the numerator and denominator.

Apply the division rule of fractions by multiplying the numerator by the reciprocal or inverse of the denominator.

Simplify, if necessary.

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2 and 2/3 x 0.40 <br> Plz help
BartSMP [9]

the mix number would be 1 and 1/15 and the improper fraction would be 16/15. 2 and 2/3 into a improper fraction and multiply it by 2/5 equals you answer.



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Christina is a dentist at the office she books 30 appointment each day three out of every 18 appointments are canceled how many
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5

Step-by-step explanation:

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Abby and Alex were practicing free throws. Abby attempted 40 shots and made 36 of them. Alex attempted 60 shots and made 54 of t
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Abby is the answer

Step-by-step explanation:

When you calculate Abby’s free throws you should realize

40 - 36 = 4

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60 - 54 = 6

When you compare the two Abby has a higher percentage of attempts because she was more successfu!

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I am building birdhouses for all of my students for an upcoming project. If I complete 2 1/2 birdhouses in 2/3 of an hour of wor
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3 years ago
Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo
Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

u\neq 0_{R2}      

v\neq 0_{R2}

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to R^{2x2} using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]

We used the first vector ''u'' as the first column of the matrix A

We used the  second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

Det(A)=6-p

We need this determinant to be different to zero

6-p\neq 0

p\neq 6

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that p\neq 6

We can write : p ∈ IR - {6}

Notice that is p=6 ⇒

u=(3,6)

v=(1,2)

If we write 3v=3(1,2)=(3,6)=u , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.

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3 years ago
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